What is (4x^2)-6x+y+1=0 ?

The solution to (4x^2)-6x+y+1=0 is Parabola with (h,k)=(3/4 , 5/4),p=-1/16

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(4x^2)-6x+y+1=0

(4 x^{2}) - 6 x + y + 1 = 0

(h, k) = (\frac{3}{4}, \frac{5}{4}), p = - \frac{1}{16}

Solution steps

4 x^{2} - 6 x + y + 1 = 0

Rewrite

4 x^{2} - 6 x + y + 1 = 0

in the standard form

4 (- \frac{1}{16}) (y - \frac{5}{4}) = (x - \frac{3}{4})^{2}

Therefore parabola properties are:

(h, k) = (\frac{3}{4}, \frac{5}{4}), p = - \frac{1}{16}

y^{2} = - 2 B x

(y + 9)^{2} - 2 (x - 4) = 0

x = y^{2} + 4 y + 3

x^{2} + 8 y - 16 = 0

x^{2} - 10 x + k (x) = 0

What is (4x^2)-6x+y+1=0 ?
The solution to (4x^2)-6x+y+1=0 is Parabola with (h,k)=(3/4 , 5/4),p=-1/16